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Proof step 5 from proof int2bits-leneq-pos (previous step)

0 < j ⊦ 2 ^ (n→i(# int2bits(j)) - 2) ≤ j, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];%l: ⊦ # bits = 0 ↔ bits = []; from specification list-data[bitlista]%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-03: ⊦ m < n → m ≤ n; from specification nat%num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n; from specification intnat%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%pot2-pos-01: ⊦ 1 < 2 ^ i ↔ 0 < i; from specification int-pot%num-n2i-ls-06: 0 ≤ i ⊦ (* i) ≤ n→i(n) ↔ i→n(i) ≤ n; from specification intnat%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i; from specification int-pot%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs